[Statics - Torques] Determining Coefficient of Static Friction

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SUMMARY

The discussion focuses on calculating the coefficient of static friction (μs) for a ladder scenario involving a woman climbing a ladder at an angle of Θ=60 degrees. The correct value of μs is established as 0.225, contrasting with an incorrect calculation of 0.333. Key steps included analyzing forces, creating a free-body diagram, and applying torque equations. The primary error identified was in the application of torque equations, specifically regarding the correct use of sine and cosine functions.

PREREQUISITES
  • Understanding of static friction and its coefficient
  • Knowledge of torque and its calculation
  • Familiarity with free-body diagrams
  • Basic trigonometry, particularly sine and cosine functions
NEXT STEPS
  • Review torque calculations in physics, focusing on radial distance and force application
  • Study the principles of static friction and its role in equilibrium problems
  • Practice drawing and analyzing free-body diagrams for various physical scenarios
  • Explore the impact of angle on friction and torque in ladder problems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to static friction and torque in practical applications.

erickbq
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Homework Statement


The top of a ladder of L and mass m is connected to a wall by a horizontal cable. The ladder makes an angle Θ=60 with the horizontal. A woman of mass 2m finds that as she climbs the ladder, the ladder begins to slip when she is one-third of the way up the ladder.

Find the coefficient of static friction between the ladder and the ground.

Homework Equations


(F⃗ net)x=ΣFx=0

(F⃗ net)y=ΣFy=0

Στ=0

τ = (radial distance)(F)

The Attempt at a Solution


a) Determined all relevant forces associated with ladder.
b) Plotted each force onto a free-body diagram.
c) Created a table with forces and their respective components
c-1) Determined torque by finding the radial distance from center of rotation (c.o.r.) to force.
d) Solved for μs.

My answer turns out to be incorrect. The answer to this question is 0.225 while I get 0.333... .
 

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The diagram is fine, but there seem to be some errors in your torque equation. Check that you have the sense of each torque correct and that you are using sin and cos appropriately.
 
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